CODEWITHSUNDEEP
phpdatabaseclasspdo
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Reputation: 23

PDO Prepared statement returns nothing even tho MySQL Statement does

i have been doing some php development for a project in class and i have encountered a problem.

The following function is supposed to return true when used with parameters i have entered myself, but it returns false : 

public function check_if_in($table, $condition){
    $request="SELECT *"; // Selecting one column
    $request=$request.' FROM '.$table.' WHERE '; // From the table i want to check in
        $keys = array_keys($condition);
        foreach($condition as $clé=>$val){
            if(!($clé == end($keys))){ // If it's not the last condition
                    $request = $request.$clé." = :".$clé." AND "; // add AND
            }
            else{
                $request = $request.$clé." = :".$clé.";"; // Add a semicolon
            }
        }
        try {
            $statement = $this->pdo->prepare($request); // Prepare the statement
        }
        catch (PDOException $e){
            die("Erreur array :" . $e->getMessage());
        }
        foreach($condition as $clé=>$val) {
            $statement->bindValue($clé, '%'.$val.'%'); // Binding all the parameters
        }

    try {
        $statement->execute();
    }
    catch (PDOException $e){
            die("Error :" . $e->getMessage());
    }
    if($statement->rowCount() > 0){
        return true;
    }
    else {
        return false;
    }
}

Where would seem to be the problem please ?

Upvotes: 1

Views: 54

Answers (1)

jeroen
jeroen

Reputation: 91734

Your problem seems to be a combination of the query and the variables you bind:

The query is built like this:

// You use `=` to compare values
$request = $request.$clé." = :".$clé

And you bind your variables like this:

// You use `%` characters as if it is a `LIKE`
$statement->bindValue($clé, '%'.$val.'%');
                             ^        ^ here you have a problem

You are using % signs like you would be using wildcards in a LIKE condition, but you are using =.

Now your query is looking for literal strings that are surrounded by % signs. Which (probably...) don't exist.

So either use LIKE instead of = or get rid of the % characters where you bind the variables:

$statement->bindValue($clé, $val);

Upvotes: 1

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